# Properties of a non-linear differential equation

I have some problems with the following differential equation, it looks a little bit confusing (because of notation), but please take a short look at it, it should be not too difficult.

$y'(t)=f(y(t)), t>0$ for $y\in C^1([0,\infty];\mathbb R^2)$ with $f\in C^1(\mathbb R^2; \mathbb R^2)$. $y_0$ is zero of $f:f(y_0)=0$. The real part of all eigenvalues of the matrix $df(y_0)$ is smaller or equal 0.

Now I have two things:

(i) I introduce $z:C^1([0,\infty]$ which is given by $z(t)=y(t)-y_0$. This should somehow satisfy $z'(t)-Az(t)=\xi (z(t))$, where $lim_{x->0}\frac{\xi (x)}{||x||}=0$ and $A\in GL(2,\mathbb R^2)$

but I do not see why. I tried to rewrite $z'(t)-Az(t)=y'(t)-A(y(t)-y_0)=f(y(t))-A(y(t)-y_0)$ but still do not see the relationship.

(ii) Every solution of $z'(t)-Az(t)=\xi (z(t))$ should satisfy $z(t)=\sum_{i=1}^{2}(z_i(0)z_h^{(i)}(t)+\int_{0}^{t}\xi_i(z(s))z_h^{(i)}(t-s) ds)$

with $z_h^{(i)}$ i=1,2 solutions of the equation $z'_h(t)=Az_h(t)$ with starting values $z_h^{(1)}=(1,0), z_h^{(2)}=(0,1)$

If you could tell how the solution of the differential equation looks like, it should be no problem to see that it satifies the equation above.

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Let $X(t)$ be a fundamental solution matrix of the differential equation $\dot{x}(t) = Ax(t) + h(t)$. Then general solution is $x(t) = X(t)[X^{-1}(t_0)x(t_0) + \int_{t_0}^t X^{-1}(s)h(s)ds]$. Where $X(t)$ has linearly independent solutions of $\dot{x} = Ax$ as its columns.
(i) Let A be the jacobian of $f$ at $y_0$. Then $\dot{z}(t) - Az(t) = f(y(t)) - A(y-y_0) = f(y_0) + Df(y_0) (y - y_0) + O(|y - y_0|^2) - A(y - y_0) = O(|y - y_0|^2) = O(|z|^2).$.
(ii) We have $\dot{z}(t) - Az(t) = \xi(z(t))$. Multiply both sides by inverse of fundamental matrix $Z^{-1}(t)$. Use fact that $Z^{-1}(t)$ is a solution of the adjoint equation $\dot{z}(t) = -z(t)A$. Then $Z^{-1}(t)\dot{z}(t) - Z^{-1}(t)Az(t) = Z^{-1}(t)\xi(z(t)) = \frac{d}{dt}[Z^{-1}(t)z(t)] = Z^{-1}(t)\xi(z(t))$. Which yields the variation of constants formula. Then doing the matrix multiplication should give you your answer.